Answer:
B
Explanation:
friction opposes motion
Friction typically Slows down objects
Option "B"How does friction affect speed?Friction and Speed
While this is almost true for a wide range of low speeds, as speed increases and air friction is reckoned with, it has been found that friction depends not only on speed, but also on speed squared and sometimes on higher powers of friction. speed.
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An object with an initial velocity of
0.12rad accelerates at 0.11rad over a
distance of 0.25 radians. What is the
final angular velocity of the object?
rad
S
Answer:
100rad because it angular velocity
quations (3.4) and (3.5) are equivalent expressions for Lagrange's equations.
Exercise 3.1 Using the Nielsen form, determine the equation of motion
for a mass m connected to a spring of constant k.
Exercise 3.2 Using the Nielsen form, determine the equations of motion
for a planet in orbit around the Sun. (Answer: mr - mro² = - GMm and
mrö +2mr00.)
3.2 Hamilton's principle
echanical system composed of N particles can be described by n = 3N
inande
Answer:Using the Nielsen form, determine the equation of motion for a mass m connected to a spring of constant k. Exercise 3.2 Using the
Explanation:
Why does the ball orbit the Earth when launched from the theoretical cannon of Newton?
A. it gets stuck in the atmosphere
B. it is magnetically attracted
C. it is attached by a rope to the Earth
D. it falls at the same rate the Earth curves
The ball orbit the Earth, when launched from the theoretical cannon of Newton, is option B. it is magnetically attracted.
Newton's Cannonball:Newton's cannonball was a hypothetical situation. Isaac Newton once proposed that gravity, which he believed to be a universal force, was the primary factor behind the planetary motion. In this experiment, Newton imagines projecting a stone or a cannonball onto the summit of a very tall mountain. The body should move away from Earth in the direction it was projected if there were no effects from gravity or air resistance.
Depending on the projectile's initial velocity and the gravitational force acting on it, the bullet will travel in a different direction. Low speeds result in a simple fallback to Earth. The Earth's surface causes the cannonball to deviate from its elliptical route.
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*example of previous incorrect/correct answer on similar question) Inside a vacuum tube, an electron is in the presence of a uniform electric field with a magnitude of 330 N/C.
(a) What is the magnitude of the acceleration of the electron (in m/s²)?
__________m/s²
(b) The electron is initially at rest. What is its speed (in m/s) after 8.00 ✕ 10−⁹ s?
______m/s
The final speed of the electron is 4.64 * 10^5 m/s.
What is the speed of the electron?Given that the mass of the electron is obtained as 9.11 * 10^-31 Kg, we have that the charge of the electron is 1.6 * 10^-19 C.
E= F/q
F = Eq
F = 330 N/C * 1.6 * 10^-19 C
F = 5.28 * 10^-17 N
F = ma
a = F/m = 5.28 * 10^-17 N/ 9.11 * 10^-31 Kg
a = 5.8 * 10^13 m/s^2
Using
v = u + at
u = 0 m/s because the electron was initially at rest
v = at
v = 5.8 * 10^13 * 8.00 ✕ 10−⁹ s
v = 4.64 * 10^5 m/s
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If something is a good conductor, what type of insulator is it?
1. also a good insulator
2. a metal
3. it depends
4. a poor insulator
4. a poor insulator
If rest other things are kept constant or unchanged then a good conductor can be termed as a poor insulator.
Answer:
4 conductors and insulators are opposites of each other.
Explanation:
8- A bottle of water weighs 25 pounds. If the bottle weighs 5 pounds, how many gal does it contain ?
A bottle of water weighs 25 pounds. If the bottle weighs 5 pounds, then it contains 0.599132136584 gal .
A bottle weighs 25 pounds ,
we know that 1 Gallon = 8.34 Lb
so, 25 lb = 2.995660682922 gal
and if the bottle weighs 5 pounds then ,
5 lb = 0.599132136584 gal
hence, 2.4 gal less.
Define pound.
a. a unit of weight equivalent to l6 ounces avoirdupois (453.59237 grams), the fundamental unit of weight in the FPS system;
b. a unit of weight equal to 12 ounces troy or 12 ounces apothecaries' (373.2418 grams).
It is an imperial unit of mass or weight measurement.
Define gallons.
In both imperial and US customary units, the gallon is a unit of volume.
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what is barometer ???
Answer:
an instrument measuring atmospheric pressure, used especially in forecasting the weather and determining altitude.
Explanation:
hope it helps ya
how to calculate the half time
The formular for calculating half life is:
T1/2=0.693/Π
Why is glass classified the way it is?
A. The bulb lights up when the switch is closed.
B. Free electrons can carry charge.
C. Delocalized electrons create a current.
D. Not enough charge carriers.
The glass is classified the way it is because of option(b)i.e, Free electrons can carry a charge.
An amorphous solid is a glass. Glass shows all the mechanical properties of a solid even though its atomic-scale structure resembles that of a supercooled liquid. Transparency, heat resistance, pressure and breakage resistance, and chemical resistance are among glass' primary properties. 90% of all manufactured glass is soda-lime glass, making it the most prevalent and least priced type.
High electronegativity and tightly bonded electrons are found in the oxygen atoms that make up glass. This is so that glass can conduct electricity, which requires a significant amount of energy to release the electrons from the oxygen atom. As a result, the glass serves as an insulator at room temperature.
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Complete the following:
When light is incident parallel to the principal axis and then strikes a lens, ___
the light will remain parallel after refracting through the lens
the light will refract through the focal point on the opposite side of the lens
the light will not refract at all
When light is incident parallel to the principal axis and then strikes a lens, the light will refract through the focal point on the opposite side of the lens.
To find the answer, we have to know about the rules followed by drawing ray-diagram.
What are the rules obeyed by light rays?If the incident ray is parallel to the principal axis, the refracted ray will pass through the opposite side's focus.The refracted ray becomes parallel to the major axis if the incident ray passes through the focus.The refracted ray follows the same path if the incident light passes through the center of the curve.Thus, we can conclude that, when light is incident parallel to the principal axis and then strikes a lens, the light will refract through the focal point on the opposite side of the lens.
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accelerates uniformly at 2.0 ms2 for 10.0s. Calculate its final velocity
Answer:
The distance is
=
7
m
Explanation:
Apply the equation of motion
s
(
t
)
=
u
t
+
1
2
a
t
2
The initial velocity is
u
=
0
m
s
−
1
The acceleration is
a
=
2
m
s
−
2
Therefore, when
t
=
3
s
, we get
s
(
3
)
=
0
+
1
2
⋅
2
⋅
3
2
=
9
m
and when
t
=
4
s
s
(
4
)
=
0
+
1
2
⋅
2
⋅
4
2
=
16
m
Therefore,
The distance travelled in the fourth second is
d
=
s
(
4
)
−
s
(
3
)
=
16
−
9
=
7
m
explain the use of low energy x-ray?
Answer:
The low energy X-ray calibration service is intended for thin-window plane-parallel chambers required for the dosimetry of superficial X-ray beams. Low-energy X-ray therapy is also known as superficial radiation therapy. This therapy uses beams of low-energy X-rays (radiation waves) to destroy skin cancer cells while not harming healthy tissue around or under the upper layers of skin.
Consider the f(x) = cos(x-C) function shown in the figure in blue color, where 0 ≤ C ≤ 2π. What is the value of parameter C for the function in the figure?
As a reference the g(x) = cos(x) function is shown in red color, and green tick marks are drawn at integer multiples of π.
The value of parameter C for the function in the figure is 2.
What is amplitude of a wave?The amplitude of a wave is the maximum displacement of the wave. It can also be described at the maximum upward displacement of a wave curve.
f(x) = Acos(x - C)
where;
A is amplitude of the waveC is phase difference of the waveWhat is angular frequency of a wave?Angular frequency is the angular displacement of any element of the wave per unit time.
From the blue colored graph; at y = 1, x = -2 cm
1 = cos(2 - C)
(2 - C) = cos^(1)
(2 - C) = 0
C = 2
Thus, the value of parameter C for the function in the figure is 2.
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a 30kg crate is pulled up a ramp 15m long and 2m high by a constant force of 100n. the crate starts from resta nd has a velocity of 2m/s when it reaches the ramp. what is the frictional force between the crate and the ramp. use the principal of conservation energy.
The frictional force between the crate and the ramp is determined as 35.2 N.
Energy lost to frictional forceApply the principle of conservation energy to calculate the change in the energy of the crate.
Change in energy of the crate = energy loss to friction
P.Ei - K.Ef = E
mgh - ¹/₂mv² = E
where;
m is mass of the crateh is vertical height traveled by the cratev is the final velocity of the crate(30)(9.8)(2) - (0.5)(30)(2²) = E
528 J = E
Frictional force between the crate and the rampE = Fd
where;
F is the frictional forced is the distance traveled by the crateF = E/d
F = (528)/(15)
F = 35.2 N
Thus, the frictional force between the crate and the ramp is determined as 35.2 N.
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A tree limb is blown loose from a tree during a storm. As it falls, it gains
speed. Which type of energy is the tree limb gaining as it falls?
O A. Kinetic energy
B. Gravitational potential energy
O C. Nuclear energy
OD. Light energy
Answer:
B Gravitational potential energy
Explanation:
If the gravitational potential energy of an object 10 m above the ground is 50 J, what is its Ep, if it moves to 30 m above the ground?
Answer:
150 J
Explanation:
Moving 3 times higher will increase the P E x 3 = 150 J
Consider a concave spherical mirr or that has focal length f = +19.5 cm.
a) What is the distance of an object from the mirror's vertex if the image is real and has the same height as the object? Follow the sign rules.
The distance of an object from the mirror's vertex if the image is real and has the same height as the object is 39 cm.
What is concave mirror?A concave mirror has a reflective surface that is curved inward and away from the light source.
Concave mirrors reflect light inward to one focal point and it usually form real and virtual images.
Object distance of the concave mirrorApply mirrors formula as shown below;
1/f = 1/v + 1/u
where;
f is the focal length of the mirrorv is the object distanceu is the image distancewhen image height = object height, magnification = 1
u/v = 1
v = u
Substitute the given parameters and solve for the distance of the object from the mirror's vertex
1/f = 1/v + 1/v
1/f = 2/v
v = 2f
v = 2(19.5 cm)
v = 39 cm
Thus, the distance of an object from the mirror's vertex if the image is real and has the same height as the object is 39 cm.
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(refer to photo attached) Determine the electric field strength at a point 1.00 cm to the left of the middle charge shown in the figure below. (Enter the magnitude of the electric field only.) _____N/C
If a charge of −4.72 µC is placed at this point, what are the magnitude and direction of the force on it? Magnitude _______N
Direction?
- toward the left
- upward
-downward
- toward the right
The electric field strength at a point 1.00 cm to the left of the middle is -2.0 x 10⁷ N/C.
The magnitude of the force is 94.4 N and direction of the force on it towards the right.
Electric field strengthThe electric field strength at a point 1.00 cm to the left of the middle is calculated as follows;
E = kq/r²
Electric field due to first charge
E1 = (9 x 10⁹ x 6 x 10⁻⁶)/(0.02)²
E1 = 1.35 x 10⁸ N/C
Electric field due to second charge
E2 = -(9 x 10⁹ x 1.5 x 10⁻⁶)/(0.01)²
E2 = - 1.35 x 10⁸ N/C
Electric field due to third charge
E3 = - (9 x 10⁹ x 2 x 10⁻⁶)/(0.03)²
E3 = -2.0 x 10⁷ N/C
Net electric fieldE = E1 + E2 + E3
E = +1.35 x 10⁸ N/C - 1.35 x 10⁸ N/C - -2.0 x 10⁷ N/C
E = -2.0 x 10⁷ N/C
Force on the charge −4.72 µCF = Eq
F = - 2.0 x 10⁷ x -4.72 x 10⁻⁶
F = 94.4 N
Thus, the direction of the force will be towards the right.
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In thicker materials the particles can move more easily, therefore the resistance has to ____________.
A. Increase
B. Decrease
C. Stay the same
D. Not enough info
In thicker materials the particles can move more easily, therefore the resistance has to decrease.
Hence Option (b) is correct.
Resistivity, which is sometimes represented by the Greek letter rho, is quantitatively equal to a specimen's resistance R times its cross-sectional area A times its length.
ρ = R × A/ l
R = ρ × l/ A
It means the Resistance of the material is inversely proportional to the area of the material.
As the area of the material increases, the resistance decreases and vice versa.
Hence, In thicker particles, there is a greater area available and that's why resistance has to decrease so that particles can move more easily.
Hence, Option (b) is correct.
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A mass M is suspended from a spring and oscillates with a period of 0.840 s. Each complete oscillation results in an amplitude reduction of a factor of 0.96 due to a small velocity dependent frictional effect. Calculate the time it takes for the total energy of the oscillator to decrease to 0.50 of its initial value.
The energy becomes 0.50 times in 6.72 s.
Let E represent the oscillator's initial energy, Et be the energy's final value at time t, where A is its beginning amplitude, At amplitude at time t, be. as the oscillator's energy increases to 0.50 times its initial value. We can replace the oscillator's total energy for the energy at time t to obtain the amplitude as shown below.
Et=0.50E
1
k(4₂)² = (0.5) - kA²
(4₂)² = (0.5) A²
At = 0.71A
So, the amplitude of the oscillator becomes 0.71 times its initial ar
0.71A = = A(0.96)¹2
log(0.71)
log(0.96)
8.4
n=
So, the time taken for n oscillation is obtained as,
t = n (0.800 s)
= (8.4) (0.800)
= 6.72 s
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NASA wants a satellite to revolve around Earth 3 times a day. What should be the radius of its orbit if we neglect the presence of the Moon? (G = 6.67 × 10-11 N ∙ m2/kg2, Mearth = 5.97 × 1024 kg)
Answer:
Approximately [tex]2.03 \times 10^{7}\; {\rm m}[/tex].
Explanation:
Assume that the radius of this orbit is [tex]r[/tex].
Let [tex]m[/tex] denote the mass of this satellite and let [tex]M[/tex] denote the mass of the Earth. At a distance of [tex]r[/tex] from the center of the earth, the magnitude of the gravitational attraction on this satellite would be [tex]G\, m\, M / (r^{2})[/tex].
The question implies that the gravitational pull from the earth is the only significant force on this satellite. Hence, the net force on this satellite would be also [tex]G\, m\, M / (r^{2})[/tex].
The acceleration of this satellite would thus be [tex]a = (\text{net force}) / (\text{mass}) = G\, M / (r^{2})[/tex].
Let [tex]\omega[/tex] denote the angular velocity of this satellite. Since this satellite in in a circular motion, the acceleration on this satellite would need to satisfy [tex]a = \omega^{2} \, r[/tex].
In other words:
[tex]\begin{aligned} \frac{G\, M}{r^{2}} = a = \omega^{2} \, r \end{aligned}[/tex].
[tex]\begin{aligned} r &= \left(\frac{G\, M}{\omega^{2}}\right)^{1/3}\end{aligned}[/tex].
The question asks for a rotation of [tex]3\times (2\, \pi) = 6\, \pi\; {\text{rad}}[/tex] within a day, which is [tex]24 \times 3600\; {\rm s}[/tex]. The angular velocity of this satellite should be:
[tex]\begin{aligned}\omega = \frac{6\, \pi}{24 \times 3600\; {\rm s}} \end{aligned}[/tex].
Substitute this value into the expression for [tex]r[/tex] and evaluate:
[tex]\begin{aligned} r &= \left(\frac{G\, M}{\omega^{2}}\right)^{1/3} \\ &= \left(\frac{(6.67 \times 10^{-11}\; {\rm N \cdot m^{2} \cdot kg^{-2}}) \times (5.97 \times 10^{24}\; {\rm kg})}{((6\, \pi) / (24 \times 3600\; {\rm s}))^{2}}\right)^{1/3} \\ &\approx 2.03 \times 10^{7}\; {\rm m}\end{aligned}[/tex].
(Note that [tex]1\; {\rm N} = 1\; {\rm kg \cdot m \cdot s^{-2}}[/tex].)
Let's begin with a simple calculation of the weight of air using density. Find the mass of air and its weight in a living room that has a 4.3m×5.4m floor and a ceiling 2.9 m high. What are the mass and weight of an equal volume of water?
a. What volume of water would have a mass equal to the mass of air in the room?
1. The mass of the air, given the data is 85.88 Kg
2. The weight of the air is 841.624 N
3. The mass of an equal volume of water is 67338 Kg
4. The weight of an equal volume of water is 659912.4 Kg
5. The volume of water that would have a mass equal to the mass of air in the room is 8.588×10⁻² m³
How to determine the volume of the airWe shall determine the volume of the air in the room by finding the volume of the room.
Volume of air = volume of room
Volume of air = 4.3 × 5.4 × 2.9
Volume of air = 67.338 m³
1. How to determine the mass of the airVolume of air = 67.338 m³Density of air = 1.2754 kg/m³Mass of air = ?Density = mass / volume
Cross multiply
Mass = Density × volume
Mass of air = 1.2754 × 67.338
Mass of air = 85.88 Kg
2. How to determine the weight of the airMass of air (m) = 85.88 KgAcceleration due to gravity (g) = 9.8 m/s²Weight of air (W) = ?W = mg
W = 85.88 × 9.8
Weight of air = 841.624 N
3. How to determine the mass of equal volume of waterVolume of air = 67.338 m³Volume of water = Volume of air = 67.338 m³Density of water = 1000 kg/m³Mass of water = ?Density = mass / volume
Cross multiply
Mass = Density × volume
Mass of water = 1000 × 67.338
Mass of water = 67338 Kg
4. How to determine the weight of equal volume of waterMass of water = 67338 KgAcceleration due to gravity (g) = 9.8 m/s²Weight of water (W) = ?W = mg
W = 67338 × 9.8
Weight of air = 659912.4 Kg
5. How to determine the volume of water with equal mass of air Mass of air = 85.88 KgMass of water = Mass of air = 85.88 KgDensity of water = 1000 kg/m³Volume of water = ?Volume = mass / density
Volume of water = 85.88 / 1000
Volume of water = 8.588×10⁻² m³
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A thin spherical shell of mass M and radius r is allowed to roll from the edge of a hemispherical bowl of radius R = 80.0 cm. It rolls down with no slipping.
1) Find the speed of the center of mass of the spherical shell when it is at the bottom of the bowl, if r is very small.
2) Repeat part 1) if r = 10.0 cm. Moment of inertia of a thing spherical shell if 2/3Mr^2.
(a) The speed of the center of mass of the spherical shell when it is at the bottom of the bowl is 3.1 m/s when the radius is 80 cm.
(b) The speed of the center of mass of the spherical shell when it is at the bottom of the bowl is 1.1 m/s when the radius is 10 cm.
Speed of the shell at the bottom of the bowl
The speed of the shell at the bottom of the bowl is calculated by applying the principle of conservation of energy.
K.E(rot) + K.E(trans) = P.E
where;
P.E is the potential energy of the ball at the initial positionK.E(rot) is rotational kinetic energyK.E(trans) is translation kinetic energy¹/₂mv² + ¹/₂Iω² = mgh
where;
I is moment of inertia of the spherical shellh is the height of fallv is the speed at the bottomω is angular speed¹/₂mv² + ¹/₂(²/₃Mr²)(v/r)² = mgh
¹/₂v² + ¹/₂(²/₃r²)(v²/r²) = gh
¹/₂v² + ¹/₂(²/₃)(v²) = gh
¹/₂v² + ¹/₃v² = gh
⁵/₆v² = gh
v² = 6gh/5
v = √(6gh/5)
Let the vertical height from the edge of bowl to the bottom , h = R = 80 cm
v = √(6 x 9.8 x 0.8 /5)
v = 3.1 m/s
When the radius = 10 cmv = √(6gh/5)
v = √(6 x 9.8 x 0.1 /5)
v = 1.1 m/s
Thus, the speed of the center of mass of the spherical shell when it is at the bottom of the bowl is 3.1 m/s when the radius is 80 cm.
The speed of the center of mass of the spherical shell when it is at the bottom of the bowl is 1.1 m/s when the radius is 10 cm.
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which element has more than 5 valency
Answer:
Sulfur (Has six valence electrons). It has maximum valency due to belonging to VI groups of the Periodic Table.
Explanation:
The electrons found in an element's outermost atomic shell are known as valence electrons.
Sulfur, which has an atomic number of 16, has an electrical configuration of 2, 8, 6, meaning it has six electrons in its outermost shell. As a result, its valence electrons will also be six.
However, in its natural condition, sulfur exists as the S8 molecule, which has the classic chair structure where each sulfur atom is covalently connected to two other sulfur atoms. In that sense, there will be 8 valence electrons.
Consequently, the answer will be 6 if you're asking about the "sulphur atom," but 8 if you're talking about sulfur in general.
Thank you ,
Eddie
*please refer to photo* An electric field of magnitude 5.25 ✕ 10^5N/C points due south at a certain location. Find the magnitude and direction of the force on a
−7.35 C charge at this location.
Magnitude _____N
Direction?
- north
-south
-east
-west
Answer:
Approximately [tex]3.86\; {\rm N}[/tex] (given that the magnitude of this charge is [tex]-7.35\; {\rm \mu C}[/tex].)
Explanation:
If a charge of magnitude [tex]q[/tex] is placed in an electric field of magnitude [tex]E[/tex], the magnitude of the electrostatic force on that charge would be [tex]F = E\, q[/tex].
The magnitude of this charge is [tex]q = 7.35\; {\rm \mu C}[/tex]. Apply the unit conversion [tex]1\; {\rm \mu C} = 10^{-6}\; {\rm C}[/tex]:
[tex]\begin{aligned} q &= 7.35\; {\mu C} \times \frac{10^{-6}\; {\rm C}}{1\; {\mu C}} = 7.35\times 10^{-6}\; {\rm C}\end{aligned}[/tex].
An electric field of magnitude [tex]E = 5.25\times 10^{5}\; {\rm N \cdot C^{-1}}[/tex] would exert on this charge a force with a magnitude of:
[tex]\begin{aligned}F &= E\, q \\ &= 5.25 \times 10^{5}\; {\rm N \cdot C^{-1}} \times (-7.35\times 10^{-6}\; {\rm C}) \\ &\approx 3.86\; {\rm N}\end{aligned}[/tex].
Note that the electric charge in this question is negative. Hence, electrostatic force on this charge would be opposite in direction to the the electric field. Since the electric field points due south, the electrostatic force on this charge would point due north.
Complete the missing information from the circuit
From the first circuit (see attached), it is correct to state that I₄ = 40mA. See the explanation below.
What is a circuit?A path that is designed to transport electric current is that is referred to as a circuit.
How do we compute for the missing details?I = 200mA
= 20mA + 140mA + I₄
I₄ = 200 - 160
I₄ = 40mA
From the above, we can rightly posit:
R₄ = 3/(40x 10⁻³)
= 75 Ω
Hence
V₆ = V₄ - V₅
= 3v - 2v
= 1v
R₆ therefore, =
V₆/I₆
= 1/ (40 x 10 ⁻³)
= 14.28 Ω
I₃ = I₄ + I₆
= 40 + 140
= 180mA
R₃ = 9/(180 x 10⁻³)
= 50Ω
R₂ = 12/20 x (10⁻³)
= 60 Ω
Then,
Vbattery =
200 x 10⁻³ x90
= 18 volts
The above indices is used to complete the image. See Circuit Image II.
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An ultracentrifuge is spinning at a speed of 80,000 rpm. The rotor that spins with
the sample can be roughly approximated as a uniform cylinder of 10 cm radius
and 8 kg mass, spinning about its symmetry axis). In order to stop the rotor in
under 30 s from when the motor is turned off, find the minimum braking torque
that must be applied.
O-19.2 Nm
-17.2 Nm
O -15.2 Nm
O-11.2 Nm
O None of the above
D. The minimum braking torque that must be applied is -11.2 Nm.
Moment of inertia of the uniform cylinderThe moment of inertia of the uniform cylinder is calculated as follows;
I = ¹/₂MR²
where;
M is mass of the cylinderR is radius of the cylinderI = (0.5)(8)(0.1²)
I = 0.04 kgm²
Minimum braking torqueτ = -Iα
where;
α is angular accelerationα = ω/t
α = (80,000 x 2π/rev x 1 min/60s) / (30 s)
α = (80,000 x 2π)/(60 x 30)
α = 279.25 rad/s²
τ = - ( 0.04 kgm²) x (279.25)
τ = -11.2 Nm
Thus, the minimum braking torque that must be applied is -11.2 Nm.
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Why do batteries discharge more quickly in cold weather?
give me answer
who will give me answer in easy language i will give him 5 star ⭐⭐⭐
Answer:
lowering the ambient temperature causes chemical reaction to proceed more slowly,so a battery used in a low temperature produces less current than that at high temperature . As cold batteries run down they quickly reach the point where they cannot deliver enough current to keep up the demand.. hence discharge more quickly in cold weather..
The Sun subtends an angle of about 0.5∘ to us on Earth, 150 million km away.
Estimate the radius of the Sun.
Express your answer to two significant figures and include the appropriate units.
The radius of the Sun is approximately 654497.9 kilometers.
A subtended angle in geometry is an angle created by a common point (in this case, the Earth) that crosses two points of a nearby circular arc (the Sun in this case). Below is a visual illustration of the subtended angle.
The law of cosine can be used to calculate the sun's radius, which is 654497.950 kilometers in kilometers.
The Sun's radius is nearly 109 times bigger than the Earth's. The Earth's radius is 6378 kilometers. Or to put it another way, both dimensions are 1: 109 ratios.
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A 0.350-kg ice puck, moving east with a speed of 5.22 m/s , has a head-on collision with a 0.950-kg puck initially at rest. Assume that the collision is perfectly elastic.
a)What is the speed of the 0.350- kg puck after the collision?
Express your answer to three significant figures and include the appropriate units.
b)What is the direction of the velocity of the 0.350- kg puck after the collision?
c)What is the speed of the 0.950- kg puck after the collision?
Express your answer to three significant figures and include the appropriate units.
d)What is the direction of the velocity of the 0.950- kg puck after the collision?
A 0.350-kg ice puck, moving east with a speed of 5.22 m/s , has a head-on collision with a 0.950-kg puck initially at rest then
(a) The speed of the 0.350- kg puck after the collision - 2.40 m/s
(b) The direction of the velocity of the 0.350- kg puck after the collision is towards West.
c) The speed of the 0.950- kg puck after the collision is 2.82 m/s
d) The direction of the velocity of the 0.950- kg puck after the collision is towards East.
Given:Mass of ice puck, m₁ = 0.350 kg
Mass of another puck, m₂ = 0.950 kg
Velocity of ice puck, v₁ = 5.22 m/s
Velocity of another puck, v₂ = 0 m/s
[tex]v^{'}_1= ?[/tex]
[tex]v^{'}_2= ?[/tex]
[tex]m_{1} v_{1} + m_{2} v_{2 }= m_{1} v^{'} _{1} + m_{2} v^{'}_{2 }[/tex]
[tex]v_{1} - v_{2} = - ( v^{'}_1 - v^{'}_2 )\\v_{1} - v_{2} = - v^{'}_1 + v^{'}_2\\v^{'}_2 = v^{'}_1 + v_{1}\\\\\\m_{1} v_{1} + m_{2} v_{2 }= m_{1} v^{'} _{1} + m_{2} v^{'}_{1 } + m_{2} v_{1 }[/tex]
[tex](0.35)(5.22) + 0 = 0.350 v^{'} _{1} + 0.950 v^{'}_{1 }+ (0.950)(5.22)\\1.827 = 0.350v^{'} _{1} + 0.950v^{'}_{1 } + 4.959\\1.827 - 4.959 = 1.3 v^{'} _{1}\\- 3.132 = 1.3 v^{'} _{1}\\v^{'} _{1} = -2.40 m/s\\\\\\[/tex]
Therefore, the speed of the 0.350- kg puck after the collision is - 2.40 m/s.
[tex]v^{'}_2 = v^{'} + v_1[/tex]
[tex]= -2.40+5.22[/tex]
[tex]= 2.82 m/s[/tex]
Therefore, the speed of the 0.950- kg puck after the collision is 2.82 m/s.
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